Question

How many liters of 0.55 M HCl is required to neutralize 1.10 g of sodium carbonate (Na2CO3)? (molar mass of Na2CO3 = 105.99 g/mol)

2HCl + Na2CO3 → 2NaCl + H2CO3

Answer 1

• Na2CO3 + 2 HCl => 2 NaCl + H2O +CO2
• So (${106 \over 2}$ )= 53 g of Na2CO3 will be neutralised by 1000 ml of 1M HCl.
• So 0.78 g of Na2CO3 will be neutralised by = 0.78/53 x 1000 ml of 1M HCl
• =( ${0,78 \over 53}$) x (${1000 \over0,53}$ ) ml of 0.53 M HCl
• = 27.77 ml of 0.53 M HCl is required.